Problem: Find $\lim_{x\to -5}\dfrac{(x-3)(x+4)}{x+5}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{4}{5}$ (Choice B) B $0$ (Choice C) C $\dfrac{9}{5}$ (Choice D) D The limit doesn't exist
Solution: Let's try to find the limit using direct substitution. $\begin{aligned} \lim_{x\to -5}\dfrac{(x-3)(x+4)}{x+5}&=\dfrac{((-5)-3)((-5)+4)}{(-5)+5} \\\\ &=\dfrac{(-8)(-1)}{0} \\\\ &=\dfrac{8}{0} \end{aligned}$ Our expression evaluates to a nonzero number over zero. In such cases, the limit doesn't exist. In conclusion, $\lim_{x\to -5}\dfrac{(x-3)(x+4)}{x+5}$ doesn't exist.